[leetcode]Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.
For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

两次遍历即可。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode *newhead = NULL;
        ListNode *newtail = NULL;
        ListNode *tmp = head;    
        while(tmp != NULL)
        {
            if(tmp->val < x)
            {
                if(newtail == NULL)
                {
                    newtail = new ListNode(tmp->val);
                    newhead = newtail;
                }
                else
                {
                    newtail->next = new ListNode(tmp->val);
                    newtail = newtail->next;
                }
            }
            tmp = tmp->next;
        }
        
        tmp = head;
        while(tmp != NULL)
        {
            if(tmp->val >= x)
            {
                if(newtail == NULL)
                {
                    newtail = new ListNode(tmp->val);
                    newhead = newtail;
                }
                else
                {
                    newtail->next = new ListNode(tmp->val);
                    newtail = newtail->next;
                }
            }
            tmp = tmp->next;
        }
        return newhead;
    }
};